[SciPy-User] understanding machine precision

Robert Kern robert.kern@gmail....
Tue Dec 14 12:47:04 CST 2010

On Tue, Dec 14, 2010 at 12:42, Keith Goodman <kwgoodman@gmail.com> wrote:
> On Tue, Dec 14, 2010 at 9:42 AM,  <josef.pktd@gmail.com> wrote:
>> I thought that we get deterministic results, with identical machine
>> precision errors, but I get (with some random a0, b0)
>>>>> for i in range(5):
>>        x = scipy.linalg.lstsq(a0,b0)[0]
>>        x2 = scipy.linalg.lstsq(a0,b0)[0]
>>        print np.max(np.abs(x-x2))
>> 9.99200722163e-016
>> 9.99200722163e-016
>> 0.0
>> 0.0
>> 9.99200722163e-016
> I've started a couple of threads in the past on repeatability. Most of
> the discussion ends up being about ATLAS. I suggest repeating the test
> without ATLAS.

On OS X with numpy linked against the builtin Accelerate.framework
(which is based off of ATLAS), I get the same result every time.

Robert Kern

"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an underlying truth."
  -- Umberto Eco

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